The marginal distribution of $X$ is $U(0, 1)$. The conditional distribution of $Y$, given $X = x$, is $U(0, e^x)$.
Find $f(y)$, the marginal PDF of $Y$ and give its support.
This should be a simple problem, but every time I attempt it I keep getting the wrong pdf for $Y$. I end up with:
$$f(y) = 1 - \frac1e$$ on $$0 < y < \frac e{e-1}.$$
But this doesn't seem to be quite right. I'm not sure where I'm going wrong.
For the joint PDF, I obtained: $$f(x, y) = e^{-x}$$ on $$0<x<1\ and \ 0 < y < e^x $$
You're answer is fine. I'm not sure why it wouldn't make sense. I'm assuming you integrated across all x values (0 to 1) of $f$$($$y$$|$$x$$)$$f$$($$x$$)$ to get your answer. Normalising gives the range you specify.